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3 years ago

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The quantity y varies as the cube of (x+2) y=32 when x=0. Find y when x=1

1 answer:

Sauron [17]3 years ago

3 0

Answer:

$y \: \alpha \: {(x + 2)}^{3} \\ y = k {(x + 2)}^{3} \\ k \: is \: a \: constant \\ when \: x = 0 \: \: y = 32 : \\ 32 = k {(0 + 2)}^{3} \\ 32 = k(8) \\ k = 4 \\ \therefore \: y = 4 {(x + 2)}^{3} \\ when \: x \: is \: 1 : \\ y = 4 {(1 + 2)}^{3} \\ y = 4 \times 27 \\ y = 108$

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